Answer:
Option B
Explanation:
I = $\int_{-1}^{1} \left(x - \left[x\right]\right)dx$
= $\int_{-1}^{1} xdx-\int_{-1}^{1}[x]dx $
= $\left[\frac{x^{2}}{2}\right]_{-1}^{1}$ - $\left[\int_{-1}^{0}[x]dx+\int_{0}^{1}[x]dx\right]$
= $\frac{1}{2}\left[1-1\right] - \left[\int_{-1}^{0}[-1]dx+\int_{0}^{1}0dx\right]$
(if -1≤x<0,[x]=-1)
(if 0≤x<1, [x] =0)
= $0 - [-x]_{-1}^{1} - 0 = 0 -.[-0 - (-1)]= 1$
